4.05/1.91 YES 4.05/1.93 proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs 4.05/1.93 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.05/1.93 4.05/1.93 4.05/1.93 Termination of the given ITRS could be proven: 4.05/1.93 4.05/1.93 (0) ITRS 4.05/1.93 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.05/1.93 (2) IDP 4.05/1.93 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.05/1.93 (4) IDP 4.05/1.93 (5) IDPNonInfProof [SOUND, 126 ms] 4.05/1.93 (6) IDP 4.05/1.93 (7) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.05/1.93 (8) TRUE 4.05/1.93 4.05/1.93 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (0) 4.05/1.93 Obligation: 4.05/1.93 ITRS problem: 4.05/1.93 4.05/1.93 The following function symbols are pre-defined: 4.05/1.93 <<< 4.05/1.93 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.93 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.93 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.93 / ~ Div: (Integer, Integer) -> Integer 4.05/1.93 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.93 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.93 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.93 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.93 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.93 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.93 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.93 + ~ Add: (Integer, Integer) -> Integer 4.05/1.93 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.93 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.93 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.93 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.93 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.93 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.93 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.93 >>> 4.05/1.93 4.05/1.93 The TRS R consists of the following rules: 4.05/1.93 minus(x, y) -> cond(x >= y + 1, x, y) 4.05/1.93 cond(FALSE, x, y) -> 0 4.05/1.93 cond(TRUE, x, y) -> 1 + minus(x, y + 1) 4.05/1.93 The set Q consists of the following terms: 4.05/1.93 minus(x0, x1) 4.05/1.93 cond(FALSE, x0, x1) 4.05/1.93 cond(TRUE, x0, x1) 4.05/1.93 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (1) ITRStoIDPProof (EQUIVALENT) 4.05/1.93 Added dependency pairs 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (2) 4.05/1.93 Obligation: 4.05/1.93 IDP problem: 4.05/1.93 The following function symbols are pre-defined: 4.05/1.93 <<< 4.05/1.93 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.93 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.93 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.93 / ~ Div: (Integer, Integer) -> Integer 4.05/1.93 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.93 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.93 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.93 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.93 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.93 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.93 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.93 + ~ Add: (Integer, Integer) -> Integer 4.05/1.93 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.93 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.93 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.93 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.93 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.93 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.93 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.93 >>> 4.05/1.93 4.05/1.93 4.05/1.93 The following domains are used: 4.05/1.93 Integer 4.05/1.93 4.05/1.93 The ITRS R consists of the following rules: 4.05/1.93 minus(x, y) -> cond(x >= y + 1, x, y) 4.05/1.93 cond(FALSE, x, y) -> 0 4.05/1.93 cond(TRUE, x, y) -> 1 + minus(x, y + 1) 4.05/1.93 4.05/1.93 The integer pair graph contains the following rules and edges: 4.05/1.93 (0): MINUS(x[0], y[0]) -> COND(x[0] >= y[0] + 1, x[0], y[0]) 4.05/1.93 (1): COND(TRUE, x[1], y[1]) -> MINUS(x[1], y[1] + 1) 4.05/1.93 4.05/1.93 (0) -> (1), if (x[0] >= y[0] + 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.05/1.93 (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) 4.05/1.93 4.05/1.93 The set Q consists of the following terms: 4.05/1.93 minus(x0, x1) 4.05/1.93 cond(FALSE, x0, x1) 4.05/1.93 cond(TRUE, x0, x1) 4.05/1.93 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (3) UsableRulesProof (EQUIVALENT) 4.05/1.93 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (4) 4.05/1.93 Obligation: 4.05/1.93 IDP problem: 4.05/1.93 The following function symbols are pre-defined: 4.05/1.93 <<< 4.05/1.93 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.93 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.93 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.93 / ~ Div: (Integer, Integer) -> Integer 4.05/1.93 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.93 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.93 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.93 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.93 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.93 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.93 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.93 + ~ Add: (Integer, Integer) -> Integer 4.05/1.93 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.93 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.93 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.93 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.93 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.93 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.93 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.93 >>> 4.05/1.93 4.05/1.93 4.05/1.93 The following domains are used: 4.05/1.93 Integer 4.05/1.93 4.05/1.93 R is empty. 4.05/1.93 4.05/1.93 The integer pair graph contains the following rules and edges: 4.05/1.93 (0): MINUS(x[0], y[0]) -> COND(x[0] >= y[0] + 1, x[0], y[0]) 4.05/1.93 (1): COND(TRUE, x[1], y[1]) -> MINUS(x[1], y[1] + 1) 4.05/1.93 4.05/1.93 (0) -> (1), if (x[0] >= y[0] + 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.05/1.93 (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) 4.05/1.93 4.05/1.93 The set Q consists of the following terms: 4.05/1.93 minus(x0, x1) 4.05/1.93 cond(FALSE, x0, x1) 4.05/1.93 cond(TRUE, x0, x1) 4.05/1.93 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (5) IDPNonInfProof (SOUND) 4.05/1.93 Used the following options for this NonInfProof: 4.05/1.93 4.05/1.93 IDPGPoloSolver: 4.05/1.93 Range: [(-1,2)] 4.05/1.93 IsNat: false 4.05/1.93 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@5f939d54 4.05/1.93 Constraint Generator: NonInfConstraintGenerator: 4.05/1.93 PathGenerator: MetricPathGenerator: 4.05/1.93 Max Left Steps: 1 4.05/1.93 Max Right Steps: 1 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 The constraints were generated the following way: 4.05/1.93 4.05/1.93 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.05/1.93 4.05/1.93 Note that final constraints are written in bold face. 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 For Pair MINUS(x, y) -> COND(>=(x, +(y, 1)), x, y) the following chains were created: 4.05/1.93 *We consider the chain MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]), COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)) which results in the following constraint: 4.05/1.93 4.05/1.93 (1) (>=(x[0], +(y[0], 1))=TRUE & x[0]=x[1] & y[0]=y[1] ==> MINUS(x[0], y[0])_>=_NonInfC & MINUS(x[0], y[0])_>=_COND(>=(x[0], +(y[0], 1)), x[0], y[0]) & (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=)) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (1) using rule (IV) which results in the following new constraint: 4.05/1.93 4.05/1.93 (2) (>=(x[0], +(y[0], 1))=TRUE ==> MINUS(x[0], y[0])_>=_NonInfC & MINUS(x[0], y[0])_>=_COND(>=(x[0], +(y[0], 1)), x[0], y[0]) & (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=)) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.05/1.93 4.05/1.93 (3) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.05/1.93 4.05/1.93 (4) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.05/1.93 4.05/1.93 (5) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 4.05/1.93 4.05/1.93 (6) (x[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.05/1.93 4.05/1.93 (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 For Pair COND(TRUE, x, y) -> MINUS(x, +(y, 1)) the following chains were created: 4.05/1.93 *We consider the chain MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]), COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)), MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]) which results in the following constraint: 4.05/1.93 4.05/1.93 (1) (>=(x[0], +(y[0], 1))=TRUE & x[0]=x[1] & y[0]=y[1] & x[1]=x[0]1 & +(y[1], 1)=y[0]1 ==> COND(TRUE, x[1], y[1])_>=_NonInfC & COND(TRUE, x[1], y[1])_>=_MINUS(x[1], +(y[1], 1)) & (U^Increasing(MINUS(x[1], +(y[1], 1))), >=)) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 4.05/1.93 4.05/1.93 (2) (>=(x[0], +(y[0], 1))=TRUE ==> COND(TRUE, x[0], y[0])_>=_NonInfC & COND(TRUE, x[0], y[0])_>=_MINUS(x[0], +(y[0], 1)) & (U^Increasing(MINUS(x[1], +(y[1], 1))), >=)) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.05/1.93 4.05/1.93 (3) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.05/1.93 4.05/1.93 (4) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.05/1.93 4.05/1.93 (5) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 4.05/1.93 4.05/1.93 (6) (x[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.05/1.93 4.05/1.93 (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 To summarize, we get the following constraints P__>=_ for the following pairs. 4.05/1.93 4.05/1.93 *MINUS(x, y) -> COND(>=(x, +(y, 1)), x, y) 4.05/1.93 4.05/1.93 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 *COND(TRUE, x, y) -> MINUS(x, +(y, 1)) 4.05/1.93 4.05/1.93 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 4.05/1.93 The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 4.05/1.93 4.05/1.93 Using the following integer polynomial ordering the resulting constraints can be solved 4.05/1.93 4.05/1.93 Polynomial interpretation over integers[POLO]: 4.05/1.93 4.05/1.93 POL(TRUE) = [1] 4.05/1.93 POL(FALSE) = 0 4.05/1.93 POL(MINUS(x_1, x_2)) = [-1] + [-1]x_2 + x_1 4.05/1.93 POL(COND(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 4.05/1.93 POL(>=(x_1, x_2)) = 0 4.05/1.93 POL(+(x_1, x_2)) = x_1 + x_2 4.05/1.93 POL(1) = [1] 4.05/1.93 4.05/1.93 4.05/1.93 The following pairs are in P_>: 4.05/1.93 4.05/1.93 4.05/1.93 COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)) 4.05/1.93 4.05/1.93 4.05/1.93 The following pairs are in P_bound: 4.05/1.93 4.05/1.93 4.05/1.93 MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]) 4.05/1.93 COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)) 4.05/1.93 4.05/1.93 4.05/1.93 The following pairs are in P_>=: 4.05/1.93 4.05/1.93 4.05/1.93 MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]) 4.05/1.93 4.05/1.93 4.05/1.93 There are no usable rules. 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (6) 4.05/1.93 Obligation: 4.05/1.93 IDP problem: 4.05/1.93 The following function symbols are pre-defined: 4.05/1.93 <<< 4.05/1.93 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.93 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.93 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.93 / ~ Div: (Integer, Integer) -> Integer 4.05/1.93 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.93 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.93 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.93 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.93 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.93 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.93 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.93 + ~ Add: (Integer, Integer) -> Integer 4.05/1.93 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.93 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.93 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.93 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.93 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.93 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.93 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.93 >>> 4.05/1.93 4.05/1.93 4.05/1.93 The following domains are used: 4.05/1.93 Integer 4.05/1.93 4.05/1.93 R is empty. 4.05/1.93 4.05/1.93 The integer pair graph contains the following rules and edges: 4.05/1.93 (0): MINUS(x[0], y[0]) -> COND(x[0] >= y[0] + 1, x[0], y[0]) 4.05/1.93 4.05/1.93 4.05/1.93 The set Q consists of the following terms: 4.05/1.93 minus(x0, x1) 4.05/1.93 cond(FALSE, x0, x1) 4.05/1.93 cond(TRUE, x0, x1) 4.05/1.93 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (7) IDependencyGraphProof (EQUIVALENT) 4.05/1.93 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.05/1.93 ---------------------------------------- 4.05/1.93 4.05/1.93 (8) 4.05/1.93 TRUE 4.20/1.96 EOF